Trig problem


by jacy
Tags: trig
jacy
jacy is offline
#1
May11-06, 10:46 PM
P: 77
I have to calculate this without using the calculator.
cos (arctan 5/12)

So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks.
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Integral
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#2
May11-06, 10:52 PM
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Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]

then use the definition of the cos of that angle to get your answer.
jacy
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#3
May11-06, 11:23 PM
P: 77
Quote Quote by Integral
Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]

then use the definition of the cos of that angle to get your answer.
am getting 12/13, am i correct

Curious3141
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#4
May11-06, 11:30 PM
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Trig problem


Quote Quote by jacy
am getting 12/13, am i correct
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).
jacy
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#5
May12-06, 08:36 AM
P: 77
Quote Quote by Curious3141
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).

Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
Curious3141
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#6
May12-06, 08:37 AM
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Quote Quote by jacy
Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
No, trig ratios have no unit. You're dividing a length by a length, so they're dimensionless.
arunbg
arunbg is offline
#7
May13-06, 09:16 AM
P: 602
Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.

For eg, if you have to do something like cos(arctan(x))
we can proceed by taking arctan(x)=y
so x=tan(y)
[tex]\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y)
or y=arccos(\frac{1}{\sqrt{1+x^2}})[/tex]
So cos(arctan(x)) = cos(y) = [itex]\frac{1}{\sqrt{1+x^2}})[/itex]
which gives the answer.This approach works for all such problems.
No messy triangles.

Arun
jacy
jacy is offline
#8
May13-06, 10:04 PM
P: 77
Thanks everyone for helping me out.


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